We define the sequence $\left(u_{n}\right)$ as follows: for every natural integer $n$, $u_{n} = \int_{0}^{1} \frac{x^{n}}{1+x} \mathrm{~d}x$.

1. Calculate $u_{0} = \int_{0}^{1} \frac{1}{1+x} \mathrm{~d}x$.
2. a) Prove that, for every natural integer $n$, $u_{n+1} + u_{n} = \frac{1}{n+1}$. b) Deduce the exact value of $u_{1}$.
3. a) Copy and complete the algorithm below so that it displays as output the term of rank $n$ of the sequence $(u_{n})$ where $n$ is a natural integer entered as input by the user.
   

   Variables :	$i$ and $n$ are natural integers, $u$ is a real number
   Input :	Enter $n$
   Initialization :	Assign to $u$ the value ...
   Processing :	\begin{tabular}{l} For $i$ varying from 1 to... | Assign to $u$ the value . . .
   End For

   \hline & \hline Output : & Display $u$ \hline \end{tabular}
   b) Using this algorithm, the following table of values was obtained:
   

   $n$	0	1	2	3	4	5	10	50	100
   $u_{n}$	0,6931	0,3069	0,1931	0,1402	0,1098	0,0902	0,0475	0,0099	0,0050

   
   What conjectures concerning the behavior of the sequence $(u_{n})$ can be made?
4. a) Prove that the sequence $(u_{n})$ is decreasing. b) Prove that the sequence $(u_{n})$ is convergent.
5. We call $\ell$ the limit of the sequence $(u_{n})$. Prove that $\ell = 0$.

The purpose of this exercise is to study the sequence $(u_n)$ defined by the value of its first term $u_1$ and, for every natural number $n$ greater than or equal to 1, by the relation: $$u_{n+1} = (n+1)u_n - 1$$
Part A

1. Verify, by detailing the calculation, that if $u_1 = 0$ then $u_4 = -17$.
2. Copy and complete the algorithm below so that by first entering in $U$ a value of $u_1$ it calculates the terms of the sequence $(u_n)$ from $u_2$ to $u_{13}$.
   For $N$ going from 1 to 12 $$U \leftarrow$$ End For
   
3. This algorithm was executed for $u_1 = 0.7$ then for $u_1 = 0.8$.
   Here are the values obtained.
   

   For $u_1 = 0.7$	For $u_1 = 0.8$
   0.4	0.6
   0.2	0.8
   -0.2	2.2
   -2	10
   -13	59
   -92	412
   -737	3295
   -6634	29654
   -66341	296539
   -729752	3261928
   -8757025	39143135
   -113841326	508860754

   
   What appears to be the limit of this sequence if $u_1 = 0.7$? And if $u_1 = 0.8$?

Part B
We consider the sequence $(I_n)$ defined for every natural number $n$, greater than or equal to 1, by: $$I_n = \int_0^1 x^n \mathrm{e}^{1-x} \mathrm{~d}x$$ We recall that the number e is the value of the exponential function at 1, that is to say that $\mathrm{e} = \mathrm{e}^1$.

1. Prove that the function $F$ defined on the interval $[0;1]$ by $F(x) = (-1-x)\mathrm{e}^{1-x}$ is an antiderivative on the interval $[0;1]$ of the function $f$ defined on the interval $[0;1]$ by $f(x) = x\mathrm{e}^{1-x}$.
2. Deduce that $I_1 = \mathrm{e} - 2$.
3. It is admitted that, for every natural number $n$ greater than or equal to 1, we have: $$I_{n+1} = (n+1)I_n - 1.$$ Use this formula to calculate $I_2$.
4. a. Justify that, for every real number $x$ in the interval $[0;1]$ and for every natural number $n$ greater than or equal to 1, we have: $0 \leqslant x^n \mathrm{e}^{1-x} \leqslant x^n \mathrm{e}$. b. Justify that: $\int_0^1 x^n \mathrm{e} \, \mathrm{d}x = \dfrac{\mathrm{e}}{n+1}$. c. Deduce that, for every natural number $n$ greater than or equal to 1, we have: $0 \leqslant I_n \leqslant \dfrac{\mathrm{e}}{n+1}$. d. Determine $\lim_{n \rightarrow +\infty} I_n$.

Part C
In this part, we denote by $n!$ the number defined, for every natural number $n$ greater than or equal to 1, by: $1! = 1$, $2! = 2 \times 1$, and if $n \geqslant 3$: $n! = n \times (n-1) \times \ldots \times 1$. And, more generally: $(n+1)! = (n+1) \times n!$

1. Prove by induction that, for every natural number $n$ greater than or equal to 1, we have: $$u_n = n! \left(u_1 - \mathrm{e} + 2\right) + I_n$$ We recall that, for every natural number $n$ greater than or equal to 1, we have: $$u_{n+1} = (n+1)u_n - 1 \quad \text{and} \quad I_{n+1} = (n+1)I_n - 1.$$
2. It is admitted that: $\lim_{n \rightarrow +\infty} n! = +\infty$. a. Determine the limit of the sequence $(u_n)$ when $u_1 = 0.7$. b. Determine the limit of the sequence $(u_n)$ when $u_1 = 0.8$.

We consider the sequence $\left( I _ { n } \right)$ defined by $I _ { 0 } = \int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { 1 - x } \mathrm {~d} x$ and for every non-zero natural number $n$
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { x ^ { n } } { 1 - x } \mathrm {~d} x$$

1. Show that $I _ { 0 } = \ln ( 2 )$.
2. a. Calculate $I _ { 0 } - I _ { 1 }$. b. Deduce $I _ { 1 }$.
3. a. Show that, for every natural number $n , I _ { n } - I _ { n + 1 } = \frac { \left( \frac { 1 } { 2 } \right) ^ { n + 1 } } { n + 1 }$. b. Propose an algorithm to determine, for a given natural number $n$, the value of $I _ { n }$.
4. Let $n$ be a non-zero natural number.

It is admitted that if $x$ belongs to the interval $\left[ 0 ; \frac { 1 } { 2 } \right]$ then $0 \leqslant \frac { x ^ { n } } { 1 - x } \leqslant \frac { 1 } { 2 ^ { n - 1 } }$. a. Show that for every non-zero natural number $n$, $0 \leqslant I _ { n } \leqslant \frac { 1 } { 2 ^ { n } }$. b. Deduce the limit of the sequence ( $I _ { n }$ ) as $n$ tends to $+ \infty$.
5. For every non-zero natural number $n$, we set
$$S _ { n } = \frac { 1 } { 2 } + \frac { \left( \frac { 1 } { 2 } \right) ^ { 2 } } { 2 } + \frac { \left( \frac { 1 } { 2 } \right) ^ { 3 } } { 3 } + \ldots + \frac { \left( \frac { 1 } { 2 } \right) ^ { n } } { n }$$
a. Show that for every non-zero natural number $n$, $S _ { n } = I _ { 0 } - I _ { n }$. b. Determine the limit of $S _ { n }$ as $n$ tends to $+ \infty$.

Part 1
We consider the function $f$ defined on the set of real numbers $\mathbb{R}$ by: $$f(x) = \left(x^2 - 4\right)\mathrm{e}^{-x}$$ We admit that the function $f$ is differentiable on $\mathbb{R}$ and we denote $f'$ its derivative function.

1. Determine the limits of the function $f$ at $-\infty$ and at $+\infty$.
2. Justify that for all real $x$, $f'(x) = \left(-x^2 + 2x + 4\right)\mathrm{e}^{-x}$.
3. Deduce the variations of the function $f$ on $\mathbb{R}$.

Part 2
We consider the sequence $(I_n)$ defined for all natural integer $n$ by $I_n = \int_{-2}^{0} x^n \mathrm{e}^{-x}\,\mathrm{d}x$.

1. Justify that $I_0 = \mathrm{e}^2 - 1$.
2. Using integration by parts, demonstrate the equality: $$I_{n+1} = (-2)^{n+1}\mathrm{e}^2 + (n+1)I_n$$
3. Deduce the exact values of $I_1$ and $I_2$.

Part 3

1. Determine the sign on $\mathbb{R}$ of the function $f$ defined in Part 1.
2. The curve $\mathscr{C}_f$ of the function $f$ is represented in an orthonormal coordinate system $(\mathrm{O}; \vec{\imath}, \vec{\jmath})$. The domain $D$ of the shaded region is bounded by the curve $\mathscr{C}_f$, the $x$-axis and the $y$-axis. Calculate the exact value, in square units, of the area $S$ of the domain $D$.

The Euler Gamma function is defined, for all real $x > 0$, by: $$\Gamma(x) = \int_{0}^{+\infty} e^{-t} t^{x-1} dt$$ Express $\Gamma(x+1)$ in terms of $x$ and $\Gamma(x)$.

We define for all real $x > 0$ the sequence $(J_{n}(x))_{n \geqslant 0}$ by: $$J_{n}(x) = \int_{0}^{1} (1-t)^{n} t^{x-1} dt$$ Show that, for all integers $n$, $n \geqslant 0$, $$\forall x > 0, \quad J_{n+1}(x) = \frac{n+1}{x} J_{n}(x+1)$$

We denote for $x \in \mathbb{R}_{+}$ and $m \in \mathbb{R}$ $$T_{m}(x) = \int_{0}^{\infty} t^{m} e^{-\left(t^{2}+x/t\right)} dt$$
a) Let $x \in \mathbb{R}_{+}^{*}$ and $m \in \mathbb{R}$. Calculate $T_{m}(x)$ in terms of $T_{m-2}(x)$ and $T_{m-3}(x)$. For this, you may consider the quantity $$\int_{A}^{B} t^{m-1}\left(2t - x/t^{2}\right) e^{-\left(t^{2}+x/t\right)} dt$$ for $0 < A < B$.
b) Let $m \in \mathbb{R}$. Find a relation between $x T_{m}^{\prime\prime\prime}(x)$, $T_{m}^{\prime\prime}(x)$ and $T_{m}(x)$ for $x \in \mathbb{R}_{+}^{*}$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$.
Let $x > 0$ and $y > 0$. Establish that $\beta ( x + 1 , y ) = \frac { x } { x + y } \beta ( x , y )$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. We want to show that for $x > 0$ and $y > 0$, $$\beta ( x , y ) = \frac { \Gamma ( x ) \Gamma ( y ) } { \Gamma ( x + y ) }$$ which will be denoted $(\mathcal{R})$.
Explain why it suffices to show the relation $(\mathcal{R})$ for $x > 1$ and $y > 1$.

We define the function $\psi$ (called the digamma function) on $\mathbb { R } ^ { + * }$ as the derivative of $x \mapsto \ln ( \Gamma ( x ) )$. For every real $x > 0 , \psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$. We admit that $\Gamma$ satisfies, for every real $x > 0$, the relation $\Gamma ( x + 1 ) = x \Gamma ( x )$.
Show that for every real $x > 0 , \psi ( x + 1 ) - \psi ( x ) = \frac { 1 } { x }$.

Show that $\Gamma$ is well defined and that for all $y > 0 , y \Gamma ( y ) = \Gamma ( y + 1 )$. Deduce that, for all $n \in \mathbb { N } , \Gamma ( n + 1 ) = n !$.
Recall that $\Gamma : ] 0 , + \infty [ \rightarrow \mathbb { R }$ is defined by $\Gamma ( y ) = \int _ { 0 } ^ { \infty } e ^ { - t } t ^ { y - 1 } d t$ and that $\Gamma \left( \frac { 1 } { 2 } \right) = \sqrt { \pi }$.

For $x \in \mathbb{R}$, $\Gamma(x) = \int_{0}^{+\infty} t^{x-1} \mathrm{e}^{-t} \mathrm{~d}t$. For all $x \in \mathcal{D}$, express $\Gamma(x+1)$ in terms of $x$ and $\Gamma(x)$.
Deduce from this, for all $x \in \mathcal{D}$ and all $n \in \mathbb{N}^{*}$, an expression for $\Gamma(x+n)$ in terms of $x$, $n$ and $\Gamma(x)$, as well as the value of $\Gamma(n)$ for all $n \geqslant 1$.

We equip $\mathbb{R}_{n}[X]$ with the inner product defined by $$\langle P, Q \rangle = \int_{-1}^{1} P(x)Q(x)\,dx$$ For $j \in \mathbb{N}$, we define the polynomial $$P_{j}(X) = \frac{1}{2^{j} j!} \frac{d^{j}}{dX^{j}}\left[(X^{2}-1)^{j}\right]$$ By convention, $P_{0} = 1$.
We denote $$g_{j} = \int_{-1}^{1} P_{j}(x)^{2}\,dx, \quad I_{j} = \int_{-1}^{1} \left(1 - x^{2}\right)^{j}\,dx$$
(a) Establish a relation between $g_{j}$ and $I_{j}$.
(b) Find a relation between $I_{j}$ and $I_{j-1} - I_{j}$, and deduce a recurrence relation for the sequence $\left(I_{j}\right)_{j \in \mathbb{N}}$.
(c) Deduce the value of $I_{j}$, then that of $g_{j}$.

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$. Show $$\forall n \in \llbracket 2, +\infty\llbracket, \forall x \in \mathbb{R}, \quad \left(1 - \frac{4x^2}{n^2}\right) I_n(x) = \frac{n-1}{n} I_{n-2}(x) \quad \text{and} \quad \left(1 - \frac{4x^2}{n^2}\right) \frac{I_n(x)}{I_n(0)} = \frac{I_{n-2}(x)}{I_{n-2}(0)}.$$

We admit the identities: $$\lim_{a \rightarrow +\infty} \int_0^a \sin(x^2) \mathrm{d}x = \lim_{a \rightarrow +\infty} \int_0^a \cos(x^2) \mathrm{d}x = \frac{\sqrt{2\pi}}{4}$$
Show that there exist real numbers $c, c' \in \mathbb{R}$ such that, as $a \rightarrow +\infty$, $$\int_0^a \sin(x^2) \mathrm{d}x = \frac{\sqrt{2\pi}}{4} + \frac{c}{a} \cos(a^2) + \frac{c'}{a^3} \sin(a^2) + O\left(\frac{1}{a^5}\right)$$

We admit the identities: $$\lim _ { a \rightarrow + \infty } \int _ { 0 } ^ { a } \sin \left( x ^ { 2 } \right) \mathrm { d } x = \lim _ { a \rightarrow + \infty } \int _ { 0 } ^ { a } \cos \left( x ^ { 2 } \right) \mathrm { d } x = \frac { \sqrt { 2 \pi } } { 4 }$$
Show that there exist real numbers $c , c ^ { \prime } \in \mathbb { R }$ such that, as $a \rightarrow + \infty$, we have $$\int _ { 0 } ^ { a } \sin \left( x ^ { 2 } \right) \mathrm { d } x = \frac { \sqrt { 2 \pi } } { 4 } + \frac { c } { a } \cos \left( a ^ { 2 } \right) + \frac { c ^ { \prime } } { a ^ { 3 } } \sin \left( a ^ { 2 } \right) + O \left( \frac { 1 } { a ^ { 5 } } \right) .$$

We define, for all $x \in \mathbb{R}^{+*}$, $$\Gamma(x) = \int_0^{+\infty} t^{x-1} \mathrm{e}^{-t} \, \mathrm{d}t$$ Show that, for all $x \in \mathbb{R}^{+*}$, $$\Gamma(x+1) = x\Gamma(x)$$

For all $n \in \mathbb { N }$, set $\mu _ { n } = \left( X ^ { n } \mid 1 \right)$ with the inner product $$( P \mid Q ) = \frac { 2 } { \pi } \int _ { 0 } ^ { 1 } P ( 4 x ) Q ( 4 x ) \frac { \sqrt { 1 - x } } { \sqrt { x } } \mathrm { ~d} x .$$ Using integration by parts, show $$\forall n \in \mathbb { N } ^ { * } , \quad 4 \mu _ { n - 1 } - \mu _ { n } = \frac { 2 \times 4 ^ { n } } { \pi } \int _ { 0 } ^ { 1 } x ^ { n - 3 / 2 } ( 1 - x ) ^ { 3 / 2 } \mathrm { d } x = \frac { 3 } { 2 n - 1 } \mu _ { n } .$$

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Determine the domain of definition of $f$ and verify that $$\forall x \in I, (x+1)f(x) = (x+2)f(x+2)$$

Establish the recurrence relation $K _ { n } = K _ { n + 1 } + \frac { 1 } { 2 n } K _ { n }$, where $K _ { n } = \int _ { 0 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Deduce that the integral $\int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x$ converges and that
$$\int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x = \int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x .$$
You may use integration by parts.

Let $f \in \mathcal { C } _ { b } ^ { 0 } ( [ 0 , + \infty [ )$. The purpose of this question is to prove that $$\left( \int _ { 0 } ^ { + \infty } f ( x ) d x \text { converges } \right) \Rightarrow \left( \lim _ { t \rightarrow 0 ^ { + } } \mathcal { L } ( f ) ( t ) = \int _ { 0 } ^ { + \infty } f ( x ) d x \right)$$
We assume that $\int _ { 0 } ^ { + \infty } f ( x ) d x$ converges.
(a) Prove that the function $$F : x \in \left[ 0 , + \infty \left[ \mapsto \int _ { x } ^ { + \infty } f ( t ) d t \right. \right.$$ is well-defined, continuous and bounded on $\left[ 0 , + \infty \right[$, of class $\mathcal { C } ^ { 1 }$ on $] 0 , + \infty [$ and satisfies $F ^ { \prime } = - f$.
(b) Prove that $\lim _ { t \rightarrow 0 ^ { + } } \mathcal { L } ( f ) ( t ) = \int _ { 0 } ^ { + \infty } f ( x ) d x$ by integration by parts.

Let $I _ { n } = \int _ { 0 } ^ { 1 } ( 1 - x ^ { 3 } ) ^ { n } d x$, where $n \in \mathbb { N }$. Then $\frac { 3 n + 1 } { 3 n } \cdot \frac { I _ { n + 1 } } { I _ { n } }$ is equal to:
(1) 1
(2) $\frac { n } { n + 1 }$
(3) $\frac { n + 1 } { n }$
(4) $\frac { 3 n + 1 } { 3 n - 2 }$

Let $I _ { n } = \int _ { 1 } ^ { e } x ^ { 19 } ( \log | x | ) ^ { n } d x$, where $n \in N$. If (20) $I _ { 10 } = \alpha I _ { 9 } + \beta I _ { 8 }$, for natural numbers $\alpha$ and $\beta$, then $\alpha - \beta$ equal to $\_\_\_\_$ .

Let $I _ { n } ( x ) = \int _ { 0 } ^ { x } \frac { 1 } { \left( t ^ { 2 } + 5 \right) ^ { n } } d t , n = 1,2,3 , \ldots$. Then
(1) $50 I _ { 6 } - 9 I _ { 5 } = x I _ { 5 } ^ { \prime }$
(2) $50 I _ { 6 } - 11 I _ { 5 } = x I _ { 5 } ^ { \prime }$
(3) $50 I _ { 6 } - 9 I _ { 5 } = I _ { 5 } ^ { \prime }$
(4) $50 I _ { 6 } - 11 I _ { 5 } = I _ { 5 } ^ { \prime }$